Quiz #16

A 760' radius horizontal curve is located on a rural minor collector with a posted speed limit of 50 mph. Advance warning signs are to be installed due to an increase in accidents occurring at this location. It has been determined that the advisory speed for this curve is 40 mph. Determine the advance placement distance for the warning signs.

A. 200 ft

B. 670 ft

C. 885 ft

D. 100 ft

I used Table 2C-4. Guidelines for Advance Placement of Warning Signs

With posted speed of 50 mph and advisory speed of 40 mph

The advanced placement distance is 100'

Blg23,

Thanks for posting this problem

blg23 said:

A 760' radius horizontal curve is located on a rural minor collector with a posted speed limit of 50 mph. Advance warning signs are to be installed due to an increase in accidents occurring at this location. It has been determined that the advisory speed for this curve is 40 mph. Determine the advance placement distance for the warning signs.

A. 200 ft

B. 670 ft

C. 885 ft

D. 100 ft

Click to expand...

Arrggghh.... you guys..... the answer choices shall be in ascending order (to match NCEES)!

nda said:

I used Table 2C-4. Guidelines for Advance Placement of Warning Signs

With posted speed of 50 mph and advisory speed of 40 mph

The advanced placement distance is 100'

Blg23,

Thanks for posting this problem

Click to expand...

For the benefit of others, you might want to mention which source Table 2C-4 is from.

ptatohed said:

nda said:

I used Table 2C-4. Guidelines for Advance Placement of Warning Signs

With posted speed of 50 mph and advisory speed of 40 mph

The advanced placement distance is 100'

Blg23,

Thanks for posting this problem

Click to expand...

For the benefit of others, you might want to mention which source Table 2C-4 is from.

Click to expand...

Sorry, I used MUTCD 2009, Chapter 2, page 108

Haha sorry ptato.

This isn't the most difficult of problems but makes you scour the MUTCD, which I anticipate having to do a few times in the PM.

Anyone else have an answer?

D - 100 feet

Table 2C-4 MUTCD Condition B, reduction to 40 mph

Great problem, because I feel the MUTCD gets neglected in the practice sets. I had this table tabbed and highlighted already, so this must have come up before. These types of problems highlight how a very easy gimme problem can be a disaster if you are not familiar with your references.

Is this answer no good?

I am starting to get confused now by the footnotes

Nope, answer is good, was just giving others a shot at it (I will admit, for whatever reason when I first read nda's answer I thought he had the incorrect answer...long day). As stated above...

MUTCD Table 2C-4

Answer D - 100 ft

I chose this one because not only does it make you page through the MUTCD, which can be a task on its own at times but because I feel the Condition B portion of the table is poorly worded. IMO, if I covered up the distances below the header, I would interpret it as listing the mph reduction from the posted speed (if that makes sense). Once you see the distances, you will obviously (hopefully) use sound judgement to reason that the large the speed reduction gets, the longer the placement distance.

Nice work all!

nda and I were going back and forth with this one during the day ....

The footnotes can be super confusing, and I think we came up with a total stopping distanvce of 450' then subtracted the 250' to get 200 ....

Thought about this more last night, then felt that is stopping distance was meant to be applied to this table, the equation would have been given in the text....

So yeah, 100 feet.

This question is written in a good way. The answers are pretty much distractors. like John said, it is great to get more familiar with MUTCD.

Next question... You guys are great. I'm loving it.

The formula for the numbers in the MUTCD Table 2C-4 is basically this:

(1.47)(Vi)(2.5) + (Vi2 - Vf2) / [30(a/32.2)] - 250, rounded to the next highest 25-ft increment, and 100 feet minimum.

Try it, it should work for any of the table numbers.

This is consistent with note #4 ("The distance is determined by providing a 2.5 second PRT, a vehicle deceleration rate of 10 feet/second2, minus the sign legibility distance of 250 feet".) and is consistent with note #6 ("The minimum advance placement distance is listed as 100 feet to provide adequate spacing between signs".).

In this case, Vi = 50mph, Vf = 40mph, a = 10ft/s2.

(1.47)(50)(2.5) + (502 - 402) / [30(10/32.2)] - 250 = 30', use 100' min.